Trignometric functions and identities

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To solve for maximum and minimum values of trigonometric functions using calculus, take the derivative of the function and set it equal to zero. The derivative of the example function, f(x) = 10cos^2(x) - 6sin(x)cos(x) + 2sin^2(x), is complex but can be simplified to find critical points. While this method can yield maximum values, it is also necessary to evaluate the second derivative or use other techniques to determine minimum values. Completing the square may be a more straightforward approach for some problems. Understanding both methods enhances problem-solving efficiency in trigonometric functions.
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Homework Statement


How to quickly solve problems on maximum and minimum values of trig functions with help of calculus:
Ex. 10cos2x-6sinxcosx+2sin2x

Homework Equations


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The Attempt at a Solution


I know the method of simplification. But i want to do it quickly with calculus. How to do that??
 
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If you "want to do it quickly with calculus" then take the derivative, set the derivative equal to 0, and solve for x. However, the derivative is fairly complicated and I'm not sure this is "quicker" than just completing the square in the original.

Setting f(x)= 10cos^2(x)- 6sin(x)cos(x)+ 2sin^2(x) then f'(x)= -20cos(x)sin(x)- 6cos^2(x)+ 6sin^2(x)+ 4sin(x)cos(x)= -16sin(x)cos(x)- 6(sin^2(x)- cos^2(x))= 0.
 
OK. but we get maximum values from that what about minimum values?
 

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